A cosmic ray electron moves at 7.80 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.00 ✕ 10−5 T. What is the radius (in m) of the circular path the electron follows? **Problem Statement:**A cosmic ray electron moves at \(7.80 \times 10^5\) m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is \(1.00 \times 10^{-5}\) T. What is the radius (in m) of the circular path the electron follows?**[ ]** m**Explanation:**To solve this problem, we can use the formula for the radius of the circular path of a charged particle moving perpendicular to a magnetic field:\[r = \frac{mv}{qB}\]where:- \(r\) is the radius of the path,- \(m\) is the mass of the electron (\(9.11 \times 10^{-31}\) kg),- \(v\) is the velocity of the electron (\(7.80 \times 10^5\) m/s),- \(q\) is the charge of the electron (\(-1.60 \times 10^{-19}\) C),- \(B\) is the magnetic field strength (\(1.00 \times 10^{-5}\) T).Using these values, you can calculate the radius \(r\).

Speed of electron "v"= 7.80 ×10^⁶ m/s. Magnetic field "B"= 1.00×10^-⁵ T . Let radius of circular…

A cosmic ray electron moves at 7.80 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.00 ✕ 10−5 T. What is the radius (in m) of the circular path the electron follows?

A cosmic ray electron moves at 7.80 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.00 ✕ 10−5 T. What is the radius (in m) of the circular path the electron follows?

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

A cosmic ray electron moves at 7.80 ✕ 10⁶ m/s perpendicular to the Earth's magnetic field at an altitude where field strength is 1.00 ✕ 10⁻⁵ T. What is the radius (in m) of the circular path the electron follows?

**Problem Statement:**

A cosmic ray electron moves at \(7.80 \times 10^5\) m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is \(1.00 \times 10^{-5}\) T. What is the radius (in m) of the circular path the electron follows?

**[ ]** m

**Explanation:**

To solve this problem, we can use the formula for the radius of the circular path of a charged particle moving perpendicular to a magnetic field:

\[
r = \frac{mv}{qB}
\]

where:
- \(r\) is the radius of the path,
- \(m\) is the mass of the electron (\(9.11 \times 10^{-31}\) kg),
- \(v\) is the velocity of the electron (\(7.80 \times 10^5\) m/s),
- \(q\) is the charge of the electron (\(-1.60 \times 10^{-19}\) C),
- \(B\) is the magnetic field strength (\(1.00 \times 10^{-5}\) T).

Using these values, you can calculate the radius \(r\).

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